# 3f4 equalisation

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3F4 Equalisation. Dr. I. J. Wassell. Introduction. When channels are fixed, we have seen that it is possible to design optimum transmit and receive filters, subject to zero ISI In practice, this is not usually possible, Ideal filters cannot be realised - PowerPoint PPT PresentationTRANSCRIPT

3F4 Equalisation

Dr. I. J. Wassell

Introduction• When channels are fixed, we have seen that it is

possible to design optimum transmit and receive filters, subject to zero ISI

• In practice, this is not usually possible,– Ideal filters cannot be realised– The channel responses can be unknown and/or time

varying– The same transmitter may be used over many

different channels

Introduction• We can improve the situation by including an

additional filtering stage at the receiver. This is known as an equalisation filter and usually it is designed to reduce ISI to a minimum

• Equalisers may be categorised as,– Fixed- The optimal equalisation filter is calculated

for a fixed (known) received pulse shape– Adaptive- The filter is adapted continuously to the

changing characteristics of the channel

Introduction

• Equalisation may be implemented using,– Analogue filters- A traditional technique

mainly confined to fixed channels. Now superseded by,

– Digital filters- Have all usual advantage of digital systems, e.g. flexibility, reliability etc. May be either fixed or adaptive. We will consider fixed equalisers implemented as digital filters

Digital Filters

• An analogue signal x(t) is sampled at times t=nT to give a ‘digital’ signal xn

0,1,......n ),(nTxxn

• The Z-transform of xn is defined analogously to the Laplace transform of a continuous signal as,

0

)(n

nn zxzX

FIR Filter• A Finite Impulse Response (FIR) filter generates a

new digital signal yn from xn using delay, multiply and addition operations

D

Xb0

D

Xb1

D

Xb2

D

Xbq

+

xn xn-1 xn-2 xn-q

yn

q

iiinqqnnnnn bxbxbxbxbxy

022110 ........

Where bi are known as the filter coefficients and delay D is equal to the sample (symbol) period T

FIR Filter• Taking the Z transform yields,

qqbzzXbzzXbzzXbzXzY )(.....)()()()( 2

21

10

q

ii

i

bzzX

bzbzbzbzX

0

22

11

0

)(

.....)(

Where z-n may be taken to mean a delay of n sample periods• Now,

)()()( zHzXzY • Hence the transfer function H(z) is,

q

ii

ibzzXzYzH

0)()()(

IIR Filters• A recursive Infinite Impulse Response Filter

generates a new digital signal yn from the input xn as follows,

D D

Xa1

D

Xa2

D

Xap

+

xn yn-1 yn-2 yn-p

yn

+

p

iiinnppnnnnn ayxayayayxy

12211 ........

Where ai are known as the filter coefficients and delay D is equal to the sample (symbol) period T

IIR Filters• Taking Z transform yields,

ppazzYazzYazzYzXzY )(.....)()()()( 2

21

1

• Rearranging,p

pazzYazzYazzYzYzX )(.....)()()()( 22

11

p

ii

i

pp

azzY

azazazzY

1

22

11

1)(

.....1)(

• Now,

p

ii

iazzXzYzH

1

1

1)()()(

Zero-Forcing Equalisers• Suppose the received pulse in a PAM system is p(t),

which suffers ISI• This signal is sampled at times t=nT to give a digital

signal pn=p(nT)

• We wish to design a digital filter HE(z) which operates on pn to eliminate ISI

• Zero ISI implies that the filter output is only non-zero in response to pulse n at sample instant n, i.e. the filter output is the unit pulse n in response to pn

Zero-Forcing Equalisers• Note that the Z transform of n is equal to 1, so,

)(1)(

1)()(

zPzH

zHzP

E

E

• Now,......)( 2

21

10

0 zpzpzpzP

0i

ii zp

• So,

0

22

11

00

1.....

1)(

1)(

i

ii

E

zpzpzpzpzPzH

Where pi are the sample values of the isolated received pulse

Zero-Forcing Equalisers• We see that this expression has the form of an IIR

filter,

.....1)( 2

21

10

0 zpzpzp

zH E

ppazazaz

.....1

1

22

11

If,

etc.

,1

22

11

0

papa

p

That is we define the amplitude of the isolated pulse at the optimum sampling point to be unity

FIR Approximations to ZFE• IIR filters are difficult to deal with in practice

– stability is not guaranteed– adaptive methods are difficult to derive– Their recursive nature makes them prone to

numerical instability• The simplest solution is to use an FIR

approximation to the ideal response

Truncated Impulse Response

• A simple way to create an FIR approximation is simply to truncate the ideal impulse response

• However, this can give rise to significant errors in the filter response

Truncated Impulse Response

• The IIR response has the form,

.....1)( 2

21

10

0 zpzpzp

zH E

• The FIR response has the form,

q

ii

iq

q bzbzbzbzbzH0

22

11

0 .....)(

• Thus we must perform polynomial division to calculate the coefficients of H(z)

Truncated Impulse Response• Example

– The unequalised pulse response at the receiver in response to a single unit amplitude transmitted pulse at sample times k = 0, 1 and 2 is, p0= 1, p1= - 0.4 and p2= - 0.2

.....1

)(1)( 2

21

10

0 zpzpzpzP

zH E

21 2.04.011)(

zz

zH E

Now,

So in this example,

Truncated Impulse Response• Performing the polynomial division,

21 2.04.01 zz 1

54321 10944.01616.0224.036.04.01 zzzzz

21 2.04.01 zz21 2.04.0 zz

321 08.016.04.0 zzz32 08.036.0 zz

432 072.0144.036.0 zzz43 072.0244.0 zz

543 0448.00896.0224.0 zzz54 0448.01616.0 zz

654 03232.006464.01616.0 zzz65 03232.010944.0 zz

E bzbzbzbzH .....)( 22

11

0

qqbzzzzz .....1616.0224.036.04.01 4321

• Truncating to 5 terms gives FIR filter with the coefficients: 1, 0.4, 0.36, 0.224, 0.1616

Now,

Direct Zero Forcing• The FIR filter equaliser output in the time

domain is,

q

iiinqqnnnnn bpbpbpbpbpy

022110 ........

• In the time domain, the zero forcing constraint is yn= 1 for one value of n and yn= 0 otherwise

Direct Zero Forcing• This constraint implies an infinite set of

simultaneous equations corresponding to,

,....., n

• However, we only have q+1 filter coefficients, so we set up q+1 equations in q+1 unknowns and solve for the coefficients

Direct Zero Forcing-Example• The sampled received pulse in response to a

single binary ‘1’ is, otherwise 0 ,3.0,6.0,8.0 210 ppp

• Design a 3-tap FIR equaliser to make the response at n=0 equal to 1, and equal to zero for n=1 and n=2

• The FIR equaliser filter output is,

2

022110

iiinnnnn bpbpbpbpy

Direct Zero Forcing-Example

• Write out previous equation for n=0, 1 and 2,

2 08.06.03.01 008.06.00 1008.0

210

210

210

nbbbnbbbnbbb

• Solving these equations gives,234.0,9375.0,25.1 210 bbb

• The zero forcing constraint is,y0 = 1, y1 = 0, y2 = 0

Error Rates and Noise• Equalisation is designed to reduce ISI and

hence increase the eye opening• However, channel noise also passes through

the equaliser and must be handled carefully to predict performance

• The frequency response of a digital filter may be obtained by substituting,

Tjez

Error Rates and Noise

• The ideal ZFE has a response,

)(1)(zP

zH E

• So in the frequency domain,

)(1)( Tj

TjE eP

eH

• Thus at frequencies where P(ejwT) is small, large noise amplification will occur.

Error Rates and Noise

0

P()Received pulse spectrum

0

HE() = 1/P()Equaliser spectral response

• In this example the low pulse spectrum response near zero will give rise to high gain and noise enhancement by the equaliser in this region.

Error Rates and Noise• What is the mean-square value (w)2 of the

noise at the equaliser output?• Suppose the equaliser filter has impulse

response bn, (n=0,..,q).• Consider the response of the equaliser to

noise alone,

q

iinin vbw

0

Error Rates and Noise• The mean-squared value is,

q

iini

q

iininw vbvbEwE

0222

0111

22

2101 02

21 inin

q

i

q

iii vvEbb

• Assume that vn has a mean-squared value, 22

vnvE

And that vn is uncorrelated white noise. Then all the terms E[vnvm] will be zero except when m=n, so,

Error Rates and Noise

• That is, the mean square noise at the filter output is that at the input multiplied by the sum squared of the filter impulse response

q

iviw b

0

222

q

iiv b

0

22

Error Rates and Noise• Hence the worst case BER may be calculated as

follows,– Calculate the eye opening h for the equalised pulse– Calculate the mean-squared noise power– Substitute into the BER expression,

2w

w

hQ2

Error Rates and Noise- Example• Returning to the previous example, calculate the

worst case BER after equalisation if unipolar line coding with transmit levels of 1V and 0V is employed.

• The sampled received pulse in response to a single binary ‘1’ is,

otherwise 0 ,3.0,6.0,8.0 210 ppp• The direct zero forcing solution is an FIR filter

with the following coefficient values,234.0,9375.0,25.1 210 bbb

Error Rates and Noise- Example• We now need to calculate the worst case

eye opening for the equalised pulse.• To do this we need to calculate the

‘residual’ values at the output of the equaliser in response to a single received pulse, pn

• From the earlier equations the FIR filter (equaliser) output is given by,

q

iiinqqnnnnn bxbxbxbxbxy

022110 ........

Error Rates and Noise- Example• In the example, the input sequence xn is the single

pulse pn and q = 2. In this case we have,

22110

2

0

bpbpbpbpy nnni

iinn

• This direct convoulution yields,

125.18.0000 bpy0) 9375.08.0()25.16.0(10011 bpbpy

0) 234.08.0() 9375.06.0()25.13.0(2011022 bpbpbpy141.0) 234.06.0() 9375.03.0(21123 bpbpy

0702.0) 234.03.0(224 bpy• Thus the equalised pulse response is,

‘residuals’

Error Rates and Noise- Example• So, remembering that for a unipolar scheme

only ‘1’s give rise to residuals, the worst case received ‘1’ is,

1 - 0.141 = 0.859 i.e., 1 other ‘1’ contributing• The worst case ‘0’ is,

0 + 0.0702 = 0.0702 i.e., 1 other ‘1’ contributing• The minimum eye opening h is,

h = 0.859 - 0.0702 = 0.789

Error Rates and Noise- Example• To calculate the rms noise at the output of

the equaliser we utilise,

q

iivw b

0

2 Where v is the rms noise at the equaliser input

• For our example,2222

22

120

2

0

2 234.0 9375.0 25.1

vvi

ivw bbbb

vw 01.2

Showing that the noise power has been increased

Error Rates and Noise- Example• The probability of bit error is given by,

we

hQP2

• Substituting for h and w gives,

vv

e QQP

2.001.22

789.0

Error Rates and Noise- Example• Note that instead of performing the

convolution to give the equaliser output in response to a single received pulse (and hence determine the residuals), an alternative is to multiply the pulse response and equaliser response in the z domain, so

)()()( zHzPzY E

22

110

22

110

zbzbbzpzpp

Error Rates and Noise- Example 4

223

12212

0211201

011000)( zbpzbpbpzbpbpbpzbpbpbpzY

Equating this to the expansion for Y(z),4

43

32

21

10)( zyzyzyzyyzY

Which yields the same expressions for the output sample values yn obtained previously by direct convolution

Other Equalisation Methods• We have seen that with ZFEs, noise can be

amplified leading to poor BER performance• Alternative design approaches take into account

noise as well as signal propagation through the equaliser

• The Minimum Mean Squared Error (MMSE) equaliser is one such approach

MMSE Equaliser• The MMSE explicitly accounts for the

presence of noise in the system• Assuming a similar model to that used

previously, then in Z transform notation,))()()(()( zVzXzHzY E

HE(z)+X(z)

V(z)

Y(z)

Where X(z) is the Z transform of the sampled received signal xn, and V(z) is the Z transform of the noise vn

MMSE Equaliser• Ideally, the equalised output yn depends only on the

transmitted symbols ak. This is not possible owing to the random noise, hence we choose to minimise the total expected mean square error (MSE) between yn and an with respect to the equaliser HE(z), i.e.,

][ 2nn ayE

MMSE Equaliser

HE(z)xn yn

an

- E[(.)2]

For a ‘fixed’ equaliser E[(.)2] is minimised by adjusting the coefficients of HE(z). Effectively we have a trade off between noise enhancement and ISI.

MMSE equaliser formulation

From data source

minimise

MMSE Equaliser• The solution has the form,

oE NzP

zH

)(1)(

Where P(z) is the Z transform of the channel pulse response and No is the noise PSD

• Note,– The equaliser needs knowledge of the noise PSD– If No=0, the solution is the same as the ZFE– When noise is present the ZFE solution is modified to

make a trade-off between ISI and noise amplification

Non-Linear Equalisation

• The equalisers considered so far are linear, since they simply involve linear filtering operations

• An alternative we consider now is non-linear equalisation

• An example is the Decision Feedback Equaliser (DFE)

DFE

• The DFE is a non-linear filter.• Again, its purpose is to cancel ISI.• The non-linearity allows some of the noise

problems associated with linear equalisers to be overcome

DFE• The structure is,

D

Xa1

D

Xa2

D

Xap

+

+

-Slicer

pn (sampled values of one pulse)

xn na

Where ai are known as the filter coefficients (not to be confused with the transmitted symbols an) and delay D is equal to the sample (symbol) period T

Detected symbols

DFE• The DFE has almost the same structure as the

standard IIR filter based equaliser• In the following development this relationship

is demonstrated• We see that the only significant difference is

the position of the data slicer (decision block)• A minor difference is the subtract function at

the DFE input. Its only effect is to alter the sign of the DFE coefficients compared with those in the IIR filter

DFE Development

D D

Xa1

D

Xa2

D

Xap

+

xn

yn

+

Slicerna

DD

Xa1

D

Xa2

D

Xap

+

xn+

ynSlicer

na

IIR Structure

IIR

DFE Developmentxn

+yn

Slicerna

DD

Xa1

D

Xa2

D

Xap

+

xn+

yn

Slicerna

DD

Xa1

D

Xa2

D

Xap

+ DFE

IIR

DFE• The DFE is almost a standard IIR filter• For this structure we know that the ZFE

solution is,

.....1)( 2

21

10

0 zpzpzp

zH E

Where p0, p1, etc. is the sampled response at the equaliser input received in response to one transmitted symbol of unity amplitude. Because we define the amplitude of the isolated pulse at the optimum sampling point to be unity, then po = 1. Comparing with the previous ZF solution where ai = -pi, this time ai = pi owing to the subtract function at the DFE input.

• The outputs of this filter with no channel noise are unit pulses, weighted by the transmitted symbol amplitudes, an

DFE Example• The sampled response to an isolated received pulse

pn is given by,p0 = 1, p1 = 0.5, p2 = -0.25

• Design a suitable DFE• From the earlier work we see that the DFE

coefficients are given by ai = pi, so,a1 = p1 = 0.5 and a2 = p2 = -0.25

• Assuming polar data pulses the effect is to– add (subtract) 0.25 (if previous but one bit is a binary

one (zero)) to the current input value to remove its effect.– subtract (add) 0.5 (if previous bit is a binary one (zero))

to the current input value to remove its effect.• Thus the effect of the previous pulses is eliminated

DFE Example

D

X

D

Xa2= p2 = -0.25

+

+

-Slicer

xn na

a1= p1 = 0.5

1

0.5

-0.5

0 T 2T 3T nT

p(nT)

p0 = 1, p1 = 0.5, p2 = -0.25

Sampled isolated pulse

DFE• In noise, we have seen that noise amplification

occurs in the IIR filter approach• To overcome this, the decision slicer is moved

inside the filter loop in the DFE• The slicer outputs the symbol estimate which

is closest to the value at its input• With no noise, this change makes no difference

because the ISI is cancelled perfectly by the IIR filter, so the slicer input is ak anyway

ka

DFE• However, in noise, the slicer acts to ‘clean-up’ the

signal, giving a noise free decision at its output.• For example, the slicer input becomes ak+vk,

where vk is the noise value

• Provided that vk is small enough the slicer still outputs the correct decision ak

• So error free cancellation continues without problems of noise amplification

DFE-Problems• Consider what happens when vk is large enough

to cause an error in the slicer decision• The error feeds back around the loop and so the

ISI is no longer cancelled.• Often a long run or ‘burst’ of errors will then be

experienced- known as error propagation• The length of the burst is of the order of 2N bits,

where N is the number of taps in the feedback filter

Automatic Equalisers• In practical communication systems the channel

is often unknown and/or time varying• To overcome this problem so called Automatic

Equalisers are employed• Two approaches are used

– Preset equaliser: The channel is measured periodically by sending some known data. The equaliser coefficients are re-calculated and subsequent data is equalised using the new coefficients.

Automatic Equalisers–Adaptive equaliser: The coefficients are adapted continuously based on the received data. A simple approach uses the Least Mean Squares (LMS) algorithm to adjust the coefficient values based on an error criterion. This approach requires that the equaliser is initially trained, so that the coefficients are initialised with approximately the correct values

An alternative adaptation algorithm is known as Recursive Least Square (RLS). This has the advantage of faster training but at the cost of higher complexity

Summary• In this section we have seen

– That in practical systems it is difficult to arrange optimum TX and RX filters subject to zero ISI

– That additional filters (equalisers) can be added to reduce ISI and improve BER performance

– Equalisers can be implemented in an analogue or digital manner and may be fixed or adaptive

– Digital implementations are fundamentally of IIR structure but may be approximated by a truncated FIR filter

– The zero-forcing criterion may lead to noise enhancement and poor performance

– MMSE and non-linear (DFE) approaches can reduce the problem of noise enhancement